Current Divider Calculator
Enter a total current and two parallel resistors and see exactly how the current splits between the branches — the current-divider rule, worked out instantly.
One current, two branches
Give the calculator the total current and the two resistances and it returns the current through each resistor using I1 = I × R2 / (R1 + R2).
Smaller resistor, larger current
The two branch currents always add up to the total. The larger share flows through the smaller resistor — the opposite of the voltage-divider rule.
What is a current divider calculator?
Total current in, branch currents out
When two resistors sit side by side in parallel, the current arriving at the junction splits between them and recombines on the far side. A current divider calculator works out that split: enter the total current and the two resistances, and it returns how many amperes flow through each branch. The two branch currents always add back up to the total, and each branch takes a share inversely proportional to its own resistance — so the easier path (the smaller resistor) carries the bigger current. This is one of the most common building blocks in circuit analysis, used whenever current has to be shared, steered, or measured across parallel paths.
Enter the total current in amperes and the two resistances in ohms to get each branch current instantly.
Both resistors share the same voltage, so each branch current is the total scaled by the other resistor over the sum of the two.
I1 = I × R2 ÷ (R1 + R2)Notice that I1 uses R2 in the numerator, not R1 — that is what makes the smaller resistor carry the larger current. The current through R2 follows the mirror formula, I2 = I × R1 / (R1 + R2), and the two always sum back to the total current I.
Suppose 2 A enters a junction of a 100 Ω resistor (R1) and a 300 Ω resistor (R2) in parallel.
Add the resistances
R1 + R2 = 100 + 300 = 400 Ω — the denominator both branches share.
Current through R1
I1 = 2 × 300 ÷ 400 = 1.5 A — the smaller resistor takes the larger share.
Current through R2
I2 = 2 × 100 ÷ 400 = 0.5 A — and 1.5 + 0.5 = 2 A, back to the total.
The two outputs are the currents flowing down each parallel branch, and the single most useful sanity check is that they add back to the total: 1.5 A through R1 plus 0.5 A through R2 equals the 2 A that arrived. The key insight is the inverse relationship — the branch with the smaller resistance carries the larger current, because current takes the path of least resistance. Here R1 is 100 Ω and R2 is 300 Ω, a 1:3 ratio, so the currents split in the opposite 3:1 ratio: three quarters of the current through R1, one quarter through R2. That inversion is exactly what trips people up coming from the voltage-divider rule, where the larger resistor gets the larger voltage. A quick mental model: each resistor's share of the current equals the other resistor's share of the total resistance. Equal resistors split the current evenly; make one resistor ten times larger and it carries about a tenth of what the other does. The numbers scale cleanly too — double the total current and both branch currents double, while their ratio stays fixed by the resistances alone.
The two-resistor current divider is exact, but it describes a specific, idealised arrangement.
Two ideal resistors in parallel only
I1 = I × R2 / (R1 + R2) applies to exactly two purely resistive branches in parallel that share the same two nodes. For three or more branches you must use the conductance form (each branch's current is proportional to its conductance, 1/R), and for AC circuits with capacitors or inductors you need impedances rather than plain resistances. The result also assumes ideal components: real wires have a small resistance, real sources are not perfectly stiff, and tolerance on the resistor values shifts the split slightly. Keep the current in amperes and both resistances in ohms, and the branch currents will always add back up to the total you entered.